We flip a fair coin (independently) three times. Define the following two events:

A = "the number of tails is odd"

B = "the first coin comes up heads"

Are the events A and B independent?

I believe this is not independent.

My though process as follows with math.

The sample space for three flips is $2^3 = 8$


Event A = $\{ THH,HTH,TTT,HHT \}$

Event B = $\{ HTT,HHT,HHH,HTH \}$

We need to prove $P(A \cap B) = P(A)\cdot P(B)$

$P(A\cap B) = \frac{2}{8} = \frac{1}{4}$

$P(A)\cdot P(B) = \frac{4}{8} \cdot \frac{4}{8} = \frac{1}{4}$

Since $P(A \cap B) = P(A)\cdot P(B)$ then events A and B are independent.

  • 1
    $\begingroup$ Use a mathematical definition of independence to resolve questions like this one. What definition do you know? $\endgroup$
    – whuber
    Commented Apr 18, 2014 at 0:29
  • $\begingroup$ Well I used the fact that for two events to be independent, event A must not have an effect on event B. In this case, technically the number of tails is odd does not depend on where the first coin comes up heads in 3 rolls, and the first coin coming up heads doesn't depend on the number of tails is odd. Id like to change my stance on this question. It is independent between A and B $\endgroup$
    – GivenPie
    Commented Apr 18, 2014 at 0:33
  • 1
    $\begingroup$ GivenPie - That's not the mathematical definition of independence. As wuber suggests, simply use the definition to resolve the question. It involves a small amount of calculation to check the condition. I suggest the easy way to begin is to simply list all the elementary events in the sample space. $\endgroup$
    – Glen_b
    Commented Apr 18, 2014 at 0:41
  • 1
    $\begingroup$ Even if you can resolve the question without reference to the mathematical definition, doing it by using the definition is not simply make-work. You need to get used to using the mathematical definition, because often that's the only way such questions can be resolved, and if you don't know how it works, you'll be stuck. It's an important part of the tools for later work. $\endgroup$
    – Glen_b
    Commented Apr 18, 2014 at 0:49
  • 1
    $\begingroup$ Yes, that's it. $\endgroup$
    – Glen_b
    Commented Apr 18, 2014 at 1:34


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